# Classical Mechanics Problem 1: Central Potential Solution ... Classical Mechanics Problem 1: Central...

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Classical Mechanics Problem 1: Central Potential Solution

a) Integrals of motion for a central potential V (r):

Angular Momentum L = rvt = r2φ̇ Energy per unit mass E = 12

( ṙ2 + v2t

) + V (r) = 12 ṙ

2 + Veff(r)

where vt is the tangential velocity and Veff is defined as

Veff(r) = V (r) + L2

2r2

If the orbit is circular, the distance of the test body from the origin is invariant: ṙ = 0, which implies that the body is always at the equilibrium-distance:

dVeff dr

= 0 ⇒ dV dr

= L2

r3 =

v2t r

= rφ̇2

then

φ̇ = ωφ = L

r2 =

( 1 r

dV

dr

)1/2

so for the period we get

Tφ = 2π ωφ

= 2π (

1 r

dV

dr

)−1/2

b) Write the orbit as in the statement of the problem:

r(t) = r0 + ²(t) with dVeff dr

(r0) = 0 and ²2 ¿ r20.

The energy per unit mass is now E = 12 ²̇ 2 + Veff(r0 + ²), and since ² is small we may

Taylor-expand the potential as

Veff(r0 + ²) = Veff(r0) + dVeff dr

(r0)² ︸ ︷︷ ︸

=0

+ 1 2

d2Veff dr2

(r0)²2 +O(²3)

so then

E − Veff(r0) = 12 ²̇ 2 +

1 2

d2Veff dr2

(r0)²2 +O(²3) = const. In the above equation we readily recognize the equation of the simple harmonic oscil- lator with

ωr = (

d2Veff dr2

)1/2

r=r0

1

and its general solution is

²(t) =

√ E − Veff(r0)

ωr cos [ωr(t− t0)]

where t0 is an arbitrary constant.

Now return to writing ωr in terms of V (r) instead of Veff(r).

ω2r = d2Veff dr2

= d2V

dr2 +

3L2

r4 =

d2V

dr2 + 3ω2φ =

d2V

dr2 +

3 r

dV

dr

ωr = (

d2V

dr2 +

3 r

dV

dr

)1/2

r=r0

= [

1 r3

d

dr

( r3

dV

dr

)]1/2

r=r0

And the radial period is

Tr = 2π ωr

c) Stability is determined by the sign of ω2r . For stability: ω2r > 0, so

1 r3

d

dr

( r3

dV

dr

) > 0

for the Yukawa-potential

V (r) = −GM r

e−kr

so the condition is

1 r3

d

dr

( r3

dV

dr

) =

GM

r3 e−kr

[ 1 + kr − (kr)2] > 0 ⇒ [1 + kr − (kr)2] > 0

[ 1 + kr − (kr)2] =

(√ 5− 1 2

+ kr

) (√ 5 + 1 2

− kr )

> 0

which is satisfied only if

kr <

(√ 5 + 1 2

)

Therefore circular orbits are unstable for

kr >

(√ 5 + 1 2

)

¥

2

d) The outermost stable circular orbit is at

r0 =

(√ 5 + 1 2k

)

its energy per unit mass is

E = V (r0) + 1 2 (r0ωφ)2 = V (r0) +

1 2

( r dV

dr

)

r=r0

1 2

GM

r0 e−kr0(kr0 − 1) = GM

r0 e−kr0

(√ 5− 1 4

) > 0

If r0 is decreased only slightly, E > 0 still and the orbit is absolutely stable ¥ The effective potential for the Yukawa-potential has the form shown in Figure 1.

r

Veff

Figure 1: Effective potential against distance from the origin

3

Classical Mechanics Problem 2: Planar Double Pendulum Solution

l

l

q 1

q 2

a) L = T − V The moment of inertia for a uniform rod of length l and mass m is

I = 1 3 ml2 about one of the ends

and Ic =

1 12

ml2 about the rod’s center

The kinetic energy term we can decompose into three parts:

T = T1 + T2,rot + T2,trans

where T1 is the kinetic energy of the first rod, T2,trans is the translational energy of the center of mass of the second rod and T2,rot is its rotational energy about its center of mass. Then

T1 = 1 6 ml2θ̇21

T2,rot = 1 24

ml2θ̇22

and T2,trans =

1 2 m

( ẋ2c + ẏ

2 c

)

where xc and yc are the coordinates of the second rod’s center of mass, so

xc = l sin θ1 + l

2 sin θ2

yc = −l cos θ1 − l2 cos θ2

from which

ẋ2c + ẏ 2 c = l

2

[ θ̇21 +

1 4 θ̇22 + θ̇1θ̇2 (sin θ1 sin θ2 + cos θ1 cos θ2)

]

4

The potential energies are simply Vi = mgyc,i, where yc,i are the vertical coordinates of the rods’ centers of mass. Since both rods are uniform, yc,i are simply the coordinates of the centers. Thus,

V1 = −mg l2 cos θ1; V2 = −mg (

l cos θ1 + l

2 cos θ2

)

The full Lagrangian is then

L = T1 + T2,rot + T2,trans − V1 − V2

= ml2 [ 2 3 θ̇21 +

1 6 θ̇22 + θ̇1θ̇2 cos (θ1 − θ2)

] + mgl

[ 3 2

cos θ1 + 1 2

cos θ2

]

b) Expand the Langrangian from part a) for small angles. The only function we have to deal with is

cos θ = 1− 1 2 θ2 +O(θ4)

Since we are going to look for normal modes with θj = θ̂j exp (iωt), where the θ̂j ¿ 1, we immediately see that in the term θ̇1θ̇2 cos (θ1 − θ2), the θ-dependence in the cosine can be dropped, because even the first θ-dependent term gives a fourth order correction. Then the approximate Lagrangian is

L = ml2 [ 2 3 θ̇21 +

1 6 θ̇22 + θ̇1θ̇2

] −mgl

[ 3 4 θ21 +

1 4 θ22

] + const.

The Euler-Lagrange equations are

d

dt

∂L

∂θ̇j =

∂L

∂θj

so in the specific case:

4 3 θ̈1 +

1 2 θ̈2 +

g

l

( 3 2 θ1

) = 0

1 2 θ̈1 +

1 3 θ̈2 +

g

l

( 1 2 θ2

) = 0

if we now look for normal modes, as mentioned, the above set of equations takes the form [ (

4 3ω

2 − 32 gl )

1 2ω

2

1 2ω

2 (

1 3ω

2 − 12 gl )

] [ θ̂1

θ̂2

] = 0

Non-trivial solutions exist if the determinant of the matrix on the left is zero. Denoting ω2 = λg/l, we can write this condition as

( 4 3 λ− 3

2

) ( 1 3 λ− 1

2

) − λ

2

4 = 0,

5

that is 7 36

λ2 − 7 6 λ +

3 4

= 0

whose solutions are λ± = 3± 6√

7 ,

so finally

ω± = [(

3± 6√ 7

) g

l

]1/2

c) To sketch the eigenmodes, find eigenvectors of the matrix in part b).

• ω2 = λ−g/l (low-frequency mode)

θ̂2 = (

3 λ − 8

3

) θ̂1 =

1 3

( 2 √

7− 1 )

θ̂1

(2 √

7− 1)/3 > 0 and real, therefore the two pendula are in phase; • ω2 = λ+g/l (high-frequency mode)

θ̂2 = (

3 λ − 8

3

) θ̂1 =

1 3

( −2 √

7− 1 )

θ̂1

(−2√7− 1)/3 < 0 and real, therefore the two pendula are perfectly out of phase.

a b

Figure 2: The low- (a) and high-frequency (b) normal modes of the planar double pendulum.

6

Electromagnetism Problem 1 Solution

a) Normal modes are products of harmonic standing waves in the x, y and z directions. For their frequencies, we have

ω = c √

k2x + k2y + k2z = c [(πnx

a

)2 +

(πny b

)2 +

(πnz b

)2]1/2 ; nx, ny, nz ∈ Z+

Since a > b, the lowest frequency has nx = 1 and either ny = 1, nz = 0 or ny = 0, nz = 1 (note that ny = 0, nz = 0 does not satisfy the boundary condition E‖,at wall = 0). Since we are told to pick the mode with ~E ‖ ŷ, the boundary conditions require

~E(r, t) = E0ŷ sin πx

a sin

πz

b cosωt

The magnetic induction we can get from Faraday’s Law:

∂ ~B

∂t = −c

( ∇× ~E

) = x̂c

∂Ey ∂z

− ẑc∂Ey ∂x

= ( x̂cE0

π

b sin

πx

a cos

πz

b − ẑcE0 π

a cos

πx

a sin

πz

b

) cosωt

~B(r, t) = −πcE0 ω

( x̂

b sin

πx

a cos

πz

b − ẑ

a cos

πx

a sin

πz

b

) sin ωt

where the frequency ω is (by the argument above)

ω = πc (

1 a2

+ 1 b2

)1/2

b) At a boundary of media, the discontinuity in the normal component of the electric field is 4π times the surface charge density σ, so

Ey(x, 0, z) = 4πσ

σ(x, 0, z) = E0 4π

sin πx

a sin

πz

b cosωt

σ(x, b, z) = −σ(x, 0, z) and

σ(0, y, z) = σ(a, y, z) = σ(x, y, 0) = σ(x, y, b) ≡ 0 Similarly, at the boundary of media the discontinuity of the tangential component of the magnetic field is given by the surface current ~κ

n̂× ~B = 4π c

~κ

7

where n̂ is a unit vector normal to the surface, so

~κ(x, 0, z) = c2E0 4ω

( ẑ

b sin

πx

a cos

πz

b +

x̂

a cos

πx

a sin

πz

b

) sin ωt

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